MHB Solving using Chebyshev's theorem

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To solve the problem using Chebyshev's theorem, the average thiamine content is 0.260 milligrams with a standard deviation of 0.005 milligrams. For at least 35/36 of the slices, k equals 6, indicating that the values must be within 6 standard deviations of the mean. For at least 80/81 of the slices, k equals 9, meaning the values must be within 9 standard deviations of the mean. The final answers require calculating the specific range of thiamine content based on these k values. The discussion emphasizes the need for both lower and upper bounds in the final answer.
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how do you solve this problem?

Average of one slice contains 0.260 milligrams of vitamin b, with a standard deviation of 0.005 milligrams. According to Chebyshev's theorem, between what values must the thiamine content be of

a) at least 35/36 of all slices of this bread
b)at least 80/81 of all slices of this bread

- - - Updated - - -

is the answer for a) 6 and b) 9?
 
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nickar1172 said:
how do you solve this problem?

Average of one slice contains 0.260 milligrams of vitamin b, with a standard deviation of 0.005 milligrams. According to Chebyshev's theorem, between what values must the thiamine content be of

a) at least 35/36 of all slices of this bread
b)at least 80/81 of all slices of this bread

- - - Updated - - -

is the answer for a) 6 and b) 9?

Hi nickar1172! :)

The question ask "between what values", meaning you're supposed to answer each question with a lower bound and an upper bound.

From Chebyshev's theorem $\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$, I do get that $k=6$ for (a) and $k=9$ for (b).

But that's an intermediate result that does not answer the question yet...
 
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